Đáp án:

$\begin{array}{l}
1)\dfrac{{\sqrt 2  + 1}}{{\sqrt 2  - 1}}\\
 = \dfrac{{\left( {\sqrt 2  + 1} \right)\left( {\sqrt 2  + 1} \right)}}{{\left( {\sqrt 2  - 1} \right)\left( {\sqrt 2  + 1} \right)}}\\
 = \dfrac{{{{\left( {\sqrt 2  + 1} \right)}^2}}}{{2 - 1}}\\
 = 2 + 2\sqrt 2  + 1\\
 = 3 + 2\sqrt 2 \\
2)2\sqrt 2 \left( {\sqrt 3  - 2} \right) + {\left( {1 + 2\sqrt 2 } \right)^2} - 2\sqrt 6 \\
 = \left( {2\sqrt 6  - 4\sqrt 2 } \right) + 1 + 4\sqrt 2  + 8 - 2\sqrt 6 \\
 = 9\\
3)\sqrt {\dfrac{4}{{{{\left( {2 - \sqrt 5 } \right)}^2}}}}  - \sqrt {\dfrac{4}{{{{\left( {2 + \sqrt 5 } \right)}^2}}}} \\
 = \dfrac{2}{{\sqrt 5  - 2}} - \dfrac{2}{{2 + \sqrt 5 }}\\
 = \dfrac{{2\left( {2 + \sqrt 5 } \right) - 2\left( {\sqrt 5  - 2} \right)}}{{\left( {\sqrt 5  - 2} \right)\left( {2 + \sqrt 5 } \right)}}\\
 = \dfrac{{4 + 2\sqrt 5  - 2\sqrt 5  + 4}}{{5 - 4}}\\
 = 8\\
4)\sqrt {2 + \sqrt 3 }  + \sqrt {2 - \sqrt 3 } \\
 = \dfrac{1}{{\sqrt 2 }}\left( {\sqrt {4 + 2\sqrt 3 }  + \sqrt {4 - 2\sqrt 3 } } \right)\\
 = \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {{{\left( {\sqrt 3  + 1} \right)}^2}}  + \sqrt {{{\left( {\sqrt 3  - 1} \right)}^2}} } \right)\\
 = \dfrac{1}{{\sqrt 2 }}\left( {\sqrt 3  + 1 + \sqrt 3  - 1} \right)\\
 = \dfrac{1}{{\sqrt 2 }}.2\sqrt 3 \\
 = \sqrt 2 .\sqrt 3 \\
 = \sqrt 6 \\
5)\left( {1 + \dfrac{{a + \sqrt a }}{{\sqrt a  + 1}}} \right)\left( {1 - \dfrac{{a - \sqrt a }}{{\sqrt a  - 1}}} \right)\\
 = \left( {1 + \dfrac{{\sqrt a \left( {\sqrt a  + 1} \right)}}{{\sqrt a  + 1}}} \right)\left( {1 - \dfrac{{\sqrt a \left( {\sqrt a  - 1} \right)}}{{\sqrt a  - 1}}} \right)\\
 = \left( {1 + \sqrt a } \right)\left( {1 - \sqrt a } \right)\\
 = 1 - a
\end{array}$